如何将字典分成几部分？

Question

So I'm wanting to split a dictionary up into parts of 19, This is only for one thing so it wouldn't remain or be saved it'll just be part of a program so it'll split them into pieces of 19 so if theres 28 it'll split them into two parts one 19 the other 9.

If that makes sense at all just struggling how to go about this xD example:

1- "69angel": {"uid": "u99bd5055900298f13821ad6" }, "bestbrittany": { "uid": "u80d4520e66090a088a8e73b2af" },

2- "blegh": { "uid": "u0c4afad0a1e1d3b9ffdd3db444d" }, "cassie84": { "uid": "u9b53c15bfd2f0e5a3741be1297" },

etc etc etc

If you wanted to spit the dict = below up into two's it'll be as so ^

dict =

    "selfbots": {
        "69angel": {"uid": "u99bd5055900298f13821ad6"
        },
        "bestbrittany": {
            "uid": "u80d4520e66090a088a8e73b2af"
        },
        "blegh": {
            "uid": "u0c4afad0a1e1d3b9ffdd3db444d"
        },
        "cassie84": {
            "uid": "u9b53c15bfd2f0e5a3741be1297"
        },
        "charlie": {
            "uid": "u983e257301e9cc6eb0f2bac49cb"
        },
        "fire4865-yy": {
            "uid": "u39f9e996dc8ca11863b539cadbc7a"
        },
        "huntress": {
            "uid": "ua2ed27b7932f647b492d8ef33c0cc"
        },
        "jerome": {
            "uid": "uf97f2811e2a2a24ad21d4e9e04565"
        },
        "kaida": {
            "uid": "ueaf35f7009d707651e32f7186bac"
        },
        "line": {
            "uid": "u7714db81cdf040de6a11caaab146"
        },
        "mickey": {
            "uid": "ub69bd9eecdaf4e643af552dd13b63"
        },
        "mrnobody": {
            "uid": "u8d322622a9400f2ef437460588ada"
        },
        "naughtyaf": {
            "uid": "u116ef7075f4bf14d0dfc4f0ba4490b"
        },
        "pinkprincess": {
            "uid": "u96f6d4900esdf22812c99bef10d9413aed0"
        },
        "queen": {
            "uid": "ub30557be34sdfdbe8d3a0be4265530f073c"
        },
        "ravenblackmoon": {
            "uid": "u78a3b5sdf1af28b029fdf6e58f886fffcc"
        },
        "sally": {
            "uid": "u3a02b2da2c6f87dsf7d1e45272cb72ce268"
        },
        "smithravi": {
            "uid": "u4d18fe936a783dsf3c36ee8d05f8409d6e6"
        },
        "sugar": {
            "uid": "uf5dedef47529a234a18dc975132d890e4af"
        },
        "twisted": {
            "uid": "u7f62d2a6baf650753234975d063316a1d8c"
        }
    },

`

Answer 1

You can decompose the dictionary into lists of keys and values:

keys, values = mydict.items()

or just work with the keys:

keys = mydict.keys()

Then you can use these to split up your dict:

dict1 = {k:mydict[k] for k in keys[:19]}
dict2 = {k:mydict[k] for k in keys[19:38]}

etc.

You could do a further list comprehension as follows:

import math

list_of_subdicts = [{k:mydict[k] for k in keys[19*i:min(len(mydict), 19*(i+1))]} for i in range(math.ceil(len(mydict)/19))]

This will give you a list of dict items, each one with 19 entries except the last one which can have up to 19 entries.

Answer 2

You want to split a dictionary into 19 parts? Well, assuming the dictionary is at least 19 entries,you could take the len() of the base dict and divide it by 19. This will be the total number of dictionaries you will need to make. It might work to make a list of dictionaries of that number? Then just loop through the dictionary and each time your iterator gets to the 19th index, switch to the next dictionary. Hope I understood your question correctly and this is some use to you. I'm not sure, just thinking of the top of my head.

Answer 3

If you don't care which dictionaries from the original list go into which group of 19, you can just use .keys() to get a list of the original dictionary keys, and add them to a new dictionary-of-dictionaries until each one gets "full" at 19, or until you run out of entries.

# Generate a large dictionary of junk
keys = [chr(c) for c in range(40,120) ]
D = dict(zip(keys,[ 'foo' for c in range(len(keys))]))

# This is the dictionary that will have all of the new "sub" dictionaries.
# For keys, just use the integers 0, 1, 2, ... .
new_dicts = { 0: {} }

# Grab each of the original dictionary entries and copy them to
# new dictionaries.  When you've filled up one with 19 characters,
# start filling a new one.  Keep going until you run out of entries.
fill_dict_number = 0
for key in D.keys():
    if len(new_dicts[fill_dict_number]) > 18:
        fill_dict_number += 1
        new_dicts[fill_dict_number] = {}
    new_dicts[fill_dict_number][key] = D[key]

print(new_dicts)